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Cross Product

Cross Product. This slideshow will be a review of the professors lectures on the Cross Product. To advance the slide, click the mouse button. To go back, right click the mouse, and choose previous. Definition.

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Cross Product

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  1. Cross Product This slideshow will be a review of the professors lectures on the Cross Product To advance the slide, click the mouse button To go back, right click the mouse, and choose previous

  2. Definition The Cross Product, unlike the dot product, yields a vector solution. Therefore, it has a magnitude and a direction. Magnitude- The magnitude of the Cross Product of vectors A and B is defined as AX B = |A| |B| sin Θ B A A Θ B

  3. Direction The direction of the Cross Product of vectors A and B is found by using the “Right Hand Rule” When the four fingers of the right hand are pointing in the direction of the first vector (i), the fingers are curled toward the second vector (j). The thumb points in the direction of the cross product direction (k).

  4. Explanation of Right Hand Rule Use the right hand rule to find i X j The four fingers are placed along the x (i) axis. The fingers are curled toward the y (j) axis. The thumb points toward the Cross Product direction, k.

  5. Cross Product Identities Since the sine of 90° = 1, and the sine of 0° = 0, from AXB = sin Θ, we can derive: iXi = 0 jXj = 0 kXk = 0 Since the angle between two i direction vectors would be 0, the equation would be: iXi = |i| |i| sin 0° = (1)(1)(0) = 0 iXj = k jXk = i kXi = j jXi = -k iXk = -j kXj = -i

  6. Cross Product Identities The Cross Product Identities can be derived by using the following alternative methods: = positive i j k i j = negative Write out i j k i j. Moving to the right yields a positive. jXk = i Moving to the left yields a negative iXk = -j Write i j k in a counterclockwise circle. Moving clockwise around the circle yields a negative. Moving counter-clockwise yields a positive.

  7. Laws of Operations • Not Commutative A X B ≠ B X A A X B = -B X A • Associative with respect to scalar multiplication a (A X B) = (a A) XB = AX (a B) • Distributive with respect to vector addition AX (B + D) = (AXB) + (AXD)

  8. Determinant Form of Cross Product Writing out the cross product A X B in component form, we can see: A X B = (Axi + Ayj + Azk) X (Bxi + Byj + Bzk) Using the distributive property, we can expand this to: = AxBx(i X i) + AxBy(i X j) + AxBz(i X k) + AyBx(j X i) + AyBy(j X j) + AyBz(j X k) + AzBx(k X i) + AzBy(kX j) + AzBz(k X k) Using the Cross Product Identities, this can be reduced to: A X B = (AyBz - AzBy)i – (AxBz - AzBx)j + (AxBy - AyBx)k

  9. Determinant Form of Cross Product We found that the expanded form of the cross product was: A X B = (AyBz-AzBy)i – (AxBz-AzBx)j + (AxBy-AyBx)k This equation can also be written in a 3X3 determinant: i j k Ax Ay Az Bx By Bz A X B = Therefore, to find the cross product of any two Cartesian vectors (A and B), write this determinant form of the vectors.

  10. Solving the Determinant A 3X3 determinant is solved by expanding the determinant into 3 2X2 determinants. i j k Ax Ay Az Bx By Bz A X B = This is done by selecting (circling) each vector, one at a time, and crossing out it’s column and row in the determinant. The remaining 2X2 determinant is multiplied by the circled vector.

  11. Solving the Determinant i j k Ax Ay Az Bx By Bz Circling the i vector yields the first term below. The second term was found by circling the j, and the third was found by circling the k. A X B = Ay Az By Bz Ax Az Bx Bz Ax Ay Bx By = i - j + k Recall that when solving a determinant, the signs alternate (+,-,+,-,….)

  12. AyBz - AzBy Solving the Determinant We must then solve the 3 2X2 determinants. Ay Az By Bz Ax Az Bx Bz Ax Ay Bx By = i - j + k Recall that a 2X2 determinant is solved by multiplying diagonally in each direction: Ay Az By Bz ( ) i i = Note that the second term is subtracted from the first.

  13. AyBz - AzBy AxBz - AzBx AxBy - AyBx Solving the Determinant We can then solve each 2X2 determinant: Ay Az By Bz Ax Az Bx Bz Ax Ay Bx By i - j + k = ( ) i - ( ) j + ( ) k This is the same answer found by writing out and simplifying the terms *Click to see solutions

  14. i j k Ax Ay Az Bx By Bz i j Ax Ay Bx By Alternative Determinant Solution The 3X3 determinant can also be solved in a way that some consider simpler. Repeat the first two columns as shown below. The cross product will be solved by evaluating the terms along the 6 diagonal lines. Evaluating down to the right will give you positive (+) and evaluating down to the left will give you negative (-) (-) (-) (-) (+) (+) (+)

  15. i j k Ax Ay Az Bx By Bz i j Ax Ay Bx By Alternative Determinant Solution = (AyBz)i + (AzBx)j + (AxBy)k - (AyBx)k - (AzBy)i - (AxBz)j This can be simplified to: (AyBz - AzBy)i – (AxBz - AzBx)j + (AxBy – AyBx)k

  16. Example of Cross Product Given: A = 8i + 4j + 6k B = 9i – 2j – 4k First, we write out a 3X3 determinant with the given information: i j k 8 4 6 9 -2 -4 A X B =

  17. (4)(-4)i + (6)(9)j + (8)(-2)k - (4)(9)k - (6)(-2)i - (8)(-4)j Solution to Cross Product Example i j k 8 4 6 9 -2 -4 We can use either method to find the solution to this determinant. We will use the method of rewriting the first two columns. i j 8 4 9 -2 i j k 8 4 6 9 -2 -4 = = (-16 + 12)i + (54 + 32)j + (-16 – 36)k *Click to see solution = -4i + 86j -52k

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